Totally disconnected space is 0-dimensional Let $X\neq \emptyset$ be a compact, perfect metrizable space. (This is a segment of an exercise, I do not know if every property will be involved in the proof)
I want to proof that $X$ is totally disconnected implies that $X$ is $0$-dimensional.
For that I have to give a basis $\mathcal{B}$ where every element is closed and open.
I tried to study the sets in $X$ which are closed and open, to construct a basis. But until now, I was not able to give such a set, besides the trivial $\emptyset$ and $X$.
$X$ is totally disconnected. So $\{x\}$ is closed. But $\{x\}$ is not open as $X$ is perfect.
I know that the clopen sets can not be finite, as this would contradict that every singleton is not open.
I do not know if this "strategy" will eventually lead to a proof.
Do you have a hint?
Thanks in advance.
 A: You need only that $X$ is totally disconnected, locally compact, and Hausdorff. Let $x\in X$, and let $U$ be an open nbhd of $x$; we want to show that $x$ has a clopen nbhd contained in $U$. $X$ is locally compact, so $x$ has an open nbhd $V$ such that $\operatorname{cl}V\subseteq U$ and $\operatorname{cl}V$ is compact. If $V$ is clopen, we’re done.
Otherwise, the total disconnectedness of $X$ implies that for each $y\in\operatorname{bdry}V$ there is set $H_y$ such that $x\in H_y$, $y\notin H_y$, and $H_y$ is clopen in $\operatorname{cl}V$. Then each $H_y$ is closed in $X$, so $\{X\setminus H_y:y\in\operatorname{bdry}V\}$ is an open cover of $\operatorname{bdry}V$. And $\operatorname{bdry}V$ is compact, so there is a finite $F\subseteq\operatorname{bdry}V$ such that
$$\operatorname{bdry}V\subseteq\bigcup_{y\in F}(X\setminus H_y)\,.$$
Let $H=\bigcap_{y\in F}H_y$; clearly $x\in H$, and $H$ is clopen in $\operatorname{cl}V$. Moreover, $H\cap\operatorname{bdry}V=\varnothing$, so $H\subseteq V\subseteq U$. Thus, $H$ is closed in the closed set $\operatorname{cl}V$ and open in the open set $V$, so $H$ is clopen in $X$ and in particular is a clopen nbhd of $x$ contained in $U$.
